题目描述
让我们一起来玩扫雷游戏！
给你一个大小为 m x n 二维字符矩阵 board ，表示扫雷游戏的盘面，其中：
'M' 代表一个 未挖出的 地雷，'E' 代表一个 未挖出的 空方块，'B' 代表没有相邻（上，下，左，右，和所有4个对角线）地雷的 已挖出的 空白方块，- 数字（
'1' 到 '8'）表示有多少地雷与这块 已挖出的 方块相邻，
'X' 则表示一个 已挖出的 地雷。
给你一个整数数组 click ，其中 click = [clickr, clickc] 表示在所有 未挖出的 方块（'M' 或者 'E'）中的下一个点击位置（clickr 是行下标，clickc 是列下标）。
根据以下规则，返回相应位置被点击后对应的盘面：
- 如果一个地雷（
'M'）被挖出，游戏就结束了- 把它改为 'X' 。
- 如果一个 没有相邻地雷 的空方块（
'E'）被挖出，修改它为（'B'），并且所有和其相邻的 未挖出 方块都应该被递归地揭露。
- 如果一个 至少与一个地雷相邻 的空方块（
'E'）被挖出，修改它为数字（'1' 到 '8' ），表示相邻地雷的数量。
- 如果在此次点击中，若无更多方块可被揭露，则返回盘面。
示例 1：
输入：board = [["E","E","E","E","E"],["E","E","M","E","E"],["E","E","E","E","E"],["E","E","E","E","E"]], click = [3,0]
输出：[["B","1","E","1","B"],["B","1","M","1","B"],["B","1","1","1","B"],["B","B","B","B","B"]]
示例 2：
输入：board = [["B","1","E","1","B"],["B","1","M","1","B"],["B","1","1","1","B"],["B","B","B","B","B"]], click = [1,2]
输出：[["B","1","E","1","B"],["B","1","X","1","B"],["B","1","1","1","B"],["B","B","B","B","B"]]
提示：
m == board.lengthn == board[i].length1 <= m, n <= 50board[i][j] 为 'M'、'E'、'B' 或数字 '1' 到 '8' 中的一个click.length == 20 <= clickr < m0 <= clickc < nboard[clickr][clickc] 为 'M' 或 'E'
解法
方法一：DFS
我们记 $click = (i, j)$，如果 $board[i][j]$ 等于 'M'，那么直接将 $board[i][j]$ 的值改为 'X' 即可。否则，我们需要统计 $board[i][j]$ 周围的地雷数量 $cnt$，如果 $cnt$ 不为 $0$，那么将 $board[i][j]$ 的值改为 $cnt$ 的字符串形式。否则，将 $board[i][j]$ 的值改为 'B'，并且递归地搜索处理 $board[i][j]$ 周围的未挖出的方块。
时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别是二维数组 $board$ 的行数和列数。
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24 | class Solution:
def updateBoard(self, board: List[List[str]], click: List[int]) -> List[List[str]]:
def dfs(i: int, j: int):
cnt = 0
for x in range(i - 1, i + 2):
for y in range(j - 1, j + 2):
if 0 <= x < m and 0 <= y < n and board[x][y] == "M":
cnt += 1
if cnt:
board[i][j] = str(cnt)
else:
board[i][j] = "B"
for x in range(i - 1, i + 2):
for y in range(j - 1, j + 2):
if 0 <= x < m and 0 <= y < n and board[x][y] == "E":
dfs(x, y)
m, n = len(board), len(board[0])
i, j = click
if board[i][j] == "M":
board[i][j] = "X"
else:
dfs(i, j)
return board
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41 | class Solution {
private char[][] board;
private int m;
private int n;
public char[][] updateBoard(char[][] board, int[] click) {
m = board.length;
n = board[0].length;
this.board = board;
int i = click[0], j = click[1];
if (board[i][j] == 'M') {
board[i][j] = 'X';
} else {
dfs(i, j);
}
return board;
}
private void dfs(int i, int j) {
int cnt = 0;
for (int x = i - 1; x <= i + 1; ++x) {
for (int y = j - 1; y <= j + 1; ++y) {
if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'M') {
++cnt;
}
}
}
if (cnt > 0) {
board[i][j] = (char) (cnt + '0');
} else {
board[i][j] = 'B';
for (int x = i - 1; x <= i + 1; ++x) {
for (int y = j - 1; y <= j + 1; ++y) {
if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'E') {
dfs(x, y);
}
}
}
}
}
}
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37 | class Solution {
public:
vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) {
int m = board.size(), n = board[0].size();
int i = click[0], j = click[1];
function<void(int, int)> dfs = [&](int i, int j) {
int cnt = 0;
for (int x = i - 1; x <= i + 1; ++x) {
for (int y = j - 1; y <= j + 1; ++y) {
if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'M') {
++cnt;
}
}
}
if (cnt) {
board[i][j] = cnt + '0';
} else {
board[i][j] = 'B';
for (int x = i - 1; x <= i + 1; ++x) {
for (int y = j - 1; y <= j + 1; ++y) {
if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'E') {
dfs(x, y);
}
}
}
}
};
if (board[i][j] == 'M') {
board[i][j] = 'X';
} else {
dfs(i, j);
}
return board;
}
};
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35 | func updateBoard(board [][]byte, click []int) [][]byte {
m, n := len(board), len(board[0])
i, j := click[0], click[1]
var dfs func(i, j int)
dfs = func(i, j int) {
cnt := 0
for x := i - 1; x <= i+1; x++ {
for y := j - 1; y <= j+1; y++ {
if x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'M' {
cnt++
}
}
}
if cnt > 0 {
board[i][j] = byte(cnt + '0')
return
}
board[i][j] = 'B'
for x := i - 1; x <= i+1; x++ {
for y := j - 1; y <= j+1; y++ {
if x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'E' {
dfs(x, y)
}
}
}
}
if board[i][j] == 'M' {
board[i][j] = 'X'
} else {
dfs(i, j)
}
return board
}
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35 | function updateBoard(board: string[][], click: number[]): string[][] {
const m = board.length;
const n = board[0].length;
const [i, j] = click;
const dfs = (i: number, j: number) => {
let cnt = 0;
for (let x = i - 1; x <= i + 1; ++x) {
for (let y = j - 1; y <= j + 1; ++y) {
if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] === 'M') {
++cnt;
}
}
}
if (cnt > 0) {
board[i][j] = cnt.toString();
return;
}
board[i][j] = 'B';
for (let x = i - 1; x <= i + 1; ++x) {
for (let y = j - 1; y <= j + 1; ++y) {
if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] === 'E') {
dfs(x, y);
}
}
}
};
if (board[i][j] === 'M') {
board[i][j] = 'X';
} else {
dfs(i, j);
}
return board;
}
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